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Let f:[-1,oo] in [-1,oo] be a function g...

Let `f:[-1,oo] in [-1,oo]` be a function given `f(x)=(x+1)^(2)-1, x ge -1`
Statement-1: The set `[x:f(x)=f^(-1)(x)]={0,1}`
Statement-2: f is a bijection.

A

Statement 1 is ture, statement 2 is true, statement 2 is a correct explanation for statement 1.

B

Statement 1 is ture, statement 2 is true, statement 2 is not a correct explanation for statement 1.

C

Statement 1 is ture, statement 2 is false.

D

Statement 1 is false, statement 2 is true.

Text Solution

Verified by Experts

The correct Answer is:
C
There is no information about co-domain therefore `f(x)` is not necessarily onto.
Therefore statement 2 is false.
However for `x ge -1, f(x) =(x+1)^(2)-1` is one-one.
Assume that `f(x)` is onto.
Now roots of `f(x)=f^(-1)(x)` lies on line `y=x`.
So we have to find roots of `f(x)=x or (x+1)^(2)-1=x`
` :. x^(2) +x=0`
` :. x=0, -1`
Hence, statement 1 is true.
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